Question

Q1.A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 13 cm, find its breadth. Which shape encloses more area?​

Answer 1

Step-by-step explanation:

Given, A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 13 cm, find its breadth.

To find, Breadth of rectangle and the shape encloses more area.

Solution :

A wire is in the shape of a square of side 10 cm.

★ Side of square = 10cm

Focus Zone : If wire is rebent into a rectangle of length 13cm. So, it means perimeter of square is equal to perimeter of rectangle.

→ Perimeter of square = Perimeter of rectangle

→ 4 × side = 2(length × breadth)

Let the breadth be b

→ 4 × 10 = 2(13 + b)

→ 40 = 26 + 2b

→ 40 - 26 = 2b

→ 14 = 2b

→ b = 14/2

→ b = 7 cm

•°• The breadth of rectangle is 7cm

Calculate area of square and rectangle

→ Area of square

→ (side)²

→ (10)²

→ 100cm²

Area of rectangle

→ length × breadth

→ 13 × 7

→ 91 cm²

•°• Square encloses more area.

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Answer 2

☀️ Given that, a wire is in the shape of square of side 10 cm and then the same wire is rebent into a rectangular shape of length 13 cm.

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❍ Here we know that,

• Length of wire [ Perimeter ] will be same when we are changing it's shape.

• Perimeter of Square = 4 × Side

• Perimeter of Rectangle = 2( Length + Breadth )

• Area of Square = Side × Side

• Area of Rectangle = Length × Breadth

H E N C E,

Finding Perimeter of Square bent wire :-

⤳ Perimeter =   10 cm × 4

⤳  Perimeter = 40 cm

Finding Breadth of Rectangle :-

• By substituting the values in the formula of perimeter as the perimeter is same as of square.

⤳  40 cm = 2( 13 cm + Breadth )

⤳  13 cm + Breadth = 40/2

⤳  13 cm + Breadth = 20 cm

⤳  Breadth = 20 cm - 13 cm

⤳  Breadth = 7 cm

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Finding the Area of Square :-

⤳  Area = 10 cm × 10 cm

⤳ Area =  100 cm²

Finding the Area of rectangle :-

⤳  Area = 13 cm × 7 cm

⤳  Area = 91 cm²

We can see that 100 cm² > 91 cm²

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• Henceforth , The wire with Square shape encloses more area.